在TAO的指导下，极限降低到1300万

发信人: saturnV (土星五号), 信区: Mathematics
标  题: 在TAO的指导下，极限降低到1300万
发信站: BBS 未名空间站 (Sun Jun  2 13:11:27 2013, 美东)

http://sbseminar.wordpress.com/2013/05/30/i-just-cant-resist-th
TAO的算功还是刚刚的

    Actually, it looks like there is considerable room to improve the
estimation of the quantity \Sigma_2, which is currently being controlled in
terms of the very large quantity \delta_2 (which remains rather large even
after the improved estimate given previously), which may lead to a
significant improvement to k_0 even without attempting either of (a) and (b)
above. I’ll try to write up some details soon.
27. Terence Tao - June 1, 2013

    OK, the details are athttp://terrytao.files.wordpress.com/2013/06/bounds.pdf. Assuming I have not made a mistake, I believe I can now improve the quantities \kappa_1, \kappa_2 in Zhang’s analysis considerably by replacing the terms involving \delta_2 with a significantly smaller expression. More specifically, whereas Zhang takes

    \kappa_1 = \delta_1 ( 1 + \delta_2^2 + k_0 \log(1+\frac{1}{4\varpi}) ) \
binom{k_0+2l_0}{k_0}

    and

    \kappa_2 = \delta_1 (1+4\varpi) (1 +\delta_2^2 + \log(1+\frac{1}{4\varpi
}) k_0 ) \binom{k_0+2l_0+1}{k_0-1}

    one can now take

    \kappa_1 = (\delta_1 + \sum_{j=1}^{1/4\varpi} \delta_1^j \delta_{2,j}^2
+ \delta k_0 \log(1+\frac{1}{4\varpi}) ) \binom{k_0+2l_0}{k_0}

    and

    \kappa_2 = (\delta_1 (1+4\varpi) + \sum_{j=1}^{1/4\varpi} \delta_1^j (1+
4\varpi)^j \delta_{2,j}^2 + \delta (1+4\varpi) k_0 \log(1+\frac{1}{4\varpi})
) \binom{k_0+2l_0+1}{k_0-1}

    where

    \delta_{2,j} := \prod_{i=1}^j (1 + k_0 \log(1+\frac{1}{i})).

    (One can replace all factors of 1/4\varpi with 292 when \varpi = 1/1168,
which is the source of all the 292′s in formulae mentioned in previous
comments.)

    The point is that \delta_{2,j} is usually much smaller than \delta_2
except when j is close to 1/4\varpi, but in that case one gets many powers
of the extremely small quantity \delta_1 = (1+1/4\varpi)^{-k_0} (rather than
just one such power, as in Zhang’s original paper).

    This should lead to a substantial improvement in the value of k_0 –
perhaps by a factor of two or more – and thus also for the final gap bound.
(Incidentally, a tiny improvement to the previous gap: when k_0 is odd one
can delete one of the extreme primes leading to a gap of p_{30798+\lfloor
2618607/2 \rfloor-1} + p_{30798+\lfloor 2618607+1/2 \rfloor -1} = 42,342,924
, if I computed correctly.)



--

※ 修改:·saturnV 於 Jun  2 13:23:17 2013 修改本文·[FROM: 67.]
※ 来源:·WWW 未名空间站 海外: mitbbs.com 中国: mitbbs.cn·[FROM: 67.]

http://www.mitbbs.com/article_t/Mathematics/31205917.html

FT中文十大 华尔街中文十大 水木十大 小百合十大 饮水思源十大 日月光华十大